Answer:
Coefficient of friction will be 0.587
Explanation:
We have given mass of the car m = 500 kg
Distance s = 18.25 m
Initial velocity of the car u = 14.5 m/sec
As the car finally stops so final velocity v = 0 m/sec
From second equation of motion



We know that acceleration is given by



So coefficient of friction will be 0.587
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,





At time
, Induced emf is,


Therefore the magnitude of the induced emf is 10.9V
300N/25 kg= divide them for the answer
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2
Move the objects faster to get more friction.